If I have a topological space $(E,\tau)$ that is metrizable for a metric $d$, and then I look at the set of all open balls in this metric $d$, will this always form a topology? If yes, will this topology be the same $\tau$ ?

  • $\begingroup$ Joseph's answer differ (edit not anymore) $\endgroup$
    – user286485
    Jan 14, 2017 at 16:33
  • 3
    $\begingroup$ In general, the collection of open balls will not be a topology. Let's think to the case of the standard topology on $\mathbb{R}$ (which is induced by the metric $(x,y)\mapsto\vert x-y\vert$) ... However, the unit balls form a base for the topology induced by a metric (by definition !), which means that every open set is the union of a family of open balls. $\endgroup$
    – Adren
    Jan 14, 2017 at 16:35

1 Answer 1

  • Yes, the topology induced by the open balls for the metric $d$ will be a topology.

I want to specify that the topology induced by the open balls is not the set of open balls, but the set of union of open balls.

  • But no, there is no reason that this topology would be the same as $\tau$.

Take for instance $E=\mathbb R$, $d=\vert .\vert$ and $\tau=\mathfrak P(\mathbb R)$ the discrete topology where every subset is open.

You have have two different topologies.

The answer for your second question is yes if you consider that metrizable means that the metric need to be deduced from the topology.


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